Uncertainty in the position of an electron (m | vedclass

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(A) The percentage error in velocity is given as $0.001\%$.$\frac{\Delta V}{V} 	imes 100 = 0.001$$\Delta V = \frac{0.001 	imes 300}{100} = 3 	imes

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$1.92 	imes 10^{-2} \ m$

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(A) The percentage error in velocity is given as $0.001\%$.$\frac{\Delta V}{V} 	imes 100 = 0.001$$\Delta V = \frac{0.001 	imes 300}{100} = 3 	imes 10^{-3} \ ms^{-1}$According to Heisenberg's uncertainty principle:$\Delta x \cdot m \Delta V \geq \frac{h}{4 \pi}$$\Delta x = \frac{h}{4 \pi m \Delta V}$Substituting the values ($h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 9.1 	imes 10^{-31} \ kg$,$\Delta V = 3 	imes 10^{-3} \ ms^{-1}$):$\Delta x = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14159 	imes 9.1 	imes 10^{-31} 	imes 3 	imes 10^{-3}}$$\Delta x \approx 1.92 	imes 10^{-2} \ m$

Uncertainty in the position of an electron (mass $= 9.1 \times 10^{-31} \ kg$) moving with a velocity $300 \ ms^{-1}$,accurate up to $0.001\%$ will be:

Similar Questions

Uncertainty in the position of an electron (mass $= 9.1 \times 10^{-31} \ kg$) moving with a velocity $300 \ ms^{-1}$ accurate up to $0.001 \ \%$ will be $(h = 6.63 \times 10^{-34} \ J \cdot s)$.

The uncertainty in momentum of an electron is $1 \times 10^{-5} \ kg \ m/s$. The uncertainty in its position will be $(h = 6.62 \times 10^{-34} \ kg \ m^2/s)$

Write down Heisenberg's uncertainty principle.

If the uncertainty in position and momentum are equal,then the uncertainty in velocity is .......

When uncertainty in position and momentum are equal,then the uncertainty in velocity is

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